0
$\begingroup$

I am new to convergence analysis of successions of functions, but I'm quite stuck on this particular exercise. $$f_n(x)=-\arctan\left(\frac{nx}{n+x}\right)$$ is a succession of functions.
Study pointwise and uniform convergence, and determine an interval of uniform convergence.

Pointwise Convergence:
$\lim_{n\to \infty}-arctan(\frac{nx}{n+x})=\lim_{n\to ∞}-\arctan(\frac{x}{1+x/n})=-\arctan(x)$.
Uniform Convergence:
$\sup_{x\in\Bbb R}{|f_n(x)-f(x)|}=$ ?
$|-arctan(\frac{nx}{n+x})+arctan(x)|\le |-arctan(\frac{nx}{n+x})| + |arctan(x)| \le \pi$
Therefore:
$\sup_{x\in\Bbb R}{|f_n(x)-f(x)|}=\pi \Rightarrow$ No uniform convergence in $\Bbb R$.

Looking for a subset of $\Bbb R$ where $f_n(x)$ converges uniformly: $$\partial_{x}(f_n(x)-f(x))=\frac{x(2n+x)}{(1+x^2)(n^2x^2+(n+x)^2)}$$

As denominator is positive, $f_n(x)-f(x)$ is monotone decreasing for $x \in (-2n,0)$.
My questions are: how can I now define a subset of $\Bbb R$ in which $f_n(x)$ converges uniformly? Which point of view (or rule) should I use when facing a similar situation?

Thank you for your time.

$\endgroup$

1 Answer 1

Reset to default
0
$\begingroup$

Your computations are fine.

You can take, for instance, the interval $[0,1]$. In that interval, $f_n-f$ is a decreasing function which takes the value $0$ at $0$. So, the minimum is attained at $1$. In other words,$$(\forall x\in[0,1]):0\geqslant f_n(x)-f(x)\geqslant f_n(1)-f(1).$$So, since $\lim_{n\to\infty}f_n(1)-f(1)=0$, the convergence is uniform in $[0,1]$.

Of course, by the same argument, any interval $[0,1]$ (with $a>0$) will do.

$\endgroup$
2
  • $\begingroup$ Thank you, you have brought some light in my reasoning. In my case, it is correct (due to $f_n - f$ decreasing in $x \in (-2n,0)$ ) to say that there is uniform convergence in $[\varphi,0)$ with $\varphi \lt 0, \forall \varphi \in \Bbb R$ as $|f_n(\varphi)-f(\varphi)|_{n \rightarrow \infty} \rightarrow 0$ and $|f_n(0)-f(0)|_{n \rightarrow \infty} \rightarrow 0$ ? $\endgroup$
    – Vrantamar
    Commented May 10, 2019 at 19:29
  • $\begingroup$ There's a subtlety there. Note that $f_n$ is undefined at $-n$. However, if $N$ is large enough, then the sequence $(f_n)_{n\geqslant N}$ converges uniformly ion $[\varphi,0]$. $\endgroup$
    – José Carlos Santos
    Commented May 10, 2019 at 20:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .